Question

The maximum peak to peak voltage of an AM wire is 24 mV and the minimum peak to peak voltage is 8 mV. The modulation factor is

• A. (a) 10%
• A. (b) 20%
• A. (c) 25%
• A. (d) 50%

Answer 1

(d)

Sol. Here,${V\frac{24}{2}}_{\max}$ and ${V\frac{8}{2}}_{\min}$

Now,

$m = \frac { V _ { \max } - V _ { \min } } { V _ { \max } + V _ { \min } } = \frac { 12 - 4 } { 12 + 4 } = \frac { 8 } { 16 } = \frac { 1 } { 2 } = 0.5 = 50 \%$